with either of these forms exactly. It is to be observed that excise officers generally consider casks to be of the first form. The measurements that are most easily made in practice are the diameters of the end, and of the middle section, and the distance between the ends; we will call these, d, D, and k, respectively, and investigate the rule in each of the above cases. (a.) Suppose the cask to be a portion of a spheroid. Its volume V will be double that of V, in the last article .. V = 2πhb2 し 1 Also by a property of the ellipse, (See fig. 50.) 1 h2 3 a2 middle diameter add the square of the end diameter, and multiply the sum by the length of the cask, this product multiplied by .2618 gives the content of the cask." (b.) Suppose the cask to have the form of a double frustum of a cone. If v is the volume required v is clearly double of the volume ABCD, fig. 48. And hence by Article 13, Mensuration of Solids, Hence the rule "To the product of the diameters add the sum of their squares, multiply this by the length of the cask, then the whole product multiplied by .2618 gives the contents of the cask." N.B. If the measurements are made in inches, the above rules give the required contents in cubic inches; to obtain the contents in gallons we must divide by 277.274, since 277.274 cubic inches go to one gallon. COR. It is evident that v, the value given by the second rule, is less than the true contents of the cask; it is to be observed, also, that V, the value given by the former rule, is generally greater than the true value; so that the true value will lie somewhere between these two results. Hence we can easily estimate the amount of accuracy in each of the above determinations. Thus, V-v is clearly greater than the difference between the true result, and either of those given by rule, and v is less than the true value of the contents, hence the error committed by either way of of the whole. Now th ( V ) "" making the calculation cannot be so great as the ( 2D + d ) D2+ Dd + d2 D (D-d). 12 which is an expression for a limit of the part of the whole, by the approximate differs from the true value. e. g. Suppose the diameters to be 18 and 20 inches respectively, and there the error committed by calculating then n N according to either rule cannot be so much as or th (very nearly) of the whole. If the cask in question had an interior 27 length of two feet, then, by the first rule, its contents are 25.47 gallons, and, by the second, its contents are 24·56 gallons. So that the error committed by either way of gauging must be less than one gallon. SPHERICAL TRIGONOMETRY. BEFORE reading the following treatise, the student will do well to reperuse the treatise of Spherical Geometry already given (p. 251, &c). He will there find the definitions enunciated and the chief properties of spherical triangles proved which are employed as the premises from which the formulas of the following treatise are deduced. It is stated in the introduction to that treatise that the chief applications of this science are found in practical astronomy and geodesy; also it is stated on p. 256, that the side of a spherical triangle measures the angle it subtends at the centre of the sphere, and hence is spoken of as an angle ;-now it is to be observed that in practical astronomy, the measurements made by the various instruments are invariably the angles subtended at the eye of the observer by arcs of the great sphere, for instance, the altitude of a star is measured directly as an angle,—so that in these cases the radius of the sphere never enters into consideration; but in the case of measurements on the earth's surface, if we have a distance measured along a great circle in miles or yards, which is to enter into our calculations, we must determine the angle these yards or miles subtend at the earth's centre; thus if a is the length in question, r the radius of the earth, the angle, then in circular measure. If e contains no, then n° = a 180° Π a = where 0 is Further it will be observed that in case the sides of a spherical triangle are small compared with the radius of the sphere, the triangle does not differ sensibly from a plane triangle: e.g. a triangle on the earth's surface the sides of which are each about a mile long will not differ sensibly from a plane triangle, unless the measurements are made with very refined instruments; hence it is manifest that the plane triangle is the limit of a spherical triangle, and accordingly we shall find that the formulas for the solution of spherical triangles are quite analogous to those that have been already deduced for the solution of plane triangles (pp. 322, 325), and we shall see that the latter can be deduced from the former by considering the plane triangle as the limit of the spherical triangle. N.B. The following results already proved on p. 259 are very important. Let A, B, C, a, b, c, be the angles and sides of any spherical triangle, and A', B', C', a' b', c', the angles and sides of the corresponding polar triangle. Then And AaB + b′ = C + c′ = 180° A' + a = B' + b = C' + c = 180° We shall employ this notation for the angles and sides of a spherical triangle and of its polar triangle throughout the following treatise. MATHEMATICAL SCIENCES.-No. XIII. 2 D (1.) To show that the Sines of the Angles of a Spherical Triangle are proportional to the Sines of the opposite sides. Let ABC be the triangle, O the centre of the sphere, join OA, OB, OC; through A draw a plane ANP perpendicular to OB, cutting the plane AOB in AN and BOC in NP, these lines are perpendicular to OB, and the angle ANP measures the inclination of the planes, and is .. equal to the angle B of the triangle. Through A draw another plane AMP perpendicular to OC, cutting AOC in AM, COB in MP, and AMP in AP, then AM and MP are perpendicular to OC, the angle AMP is equal to the angle C of the triangle, and AP is perpendicular to the plane BOA, and .. APN and APM are each right angles. Hence N M c Fig. 1. = AN The same proof holds good of the other sides and angles. COR. 2.-If through O and AP a plane be drawn cutting the surface of the sphere As before, let O be the centre of the sphere, and ABC the triangle, join OA. OB. OC, and produce the planes AOB, BOC, COA indefinitely; at A draw a plane Apq perpendicular to OA, cutting the planes AOB, BOC, COA in Ap, pq, qA respectively, then since pA is on the plane AOB, and perpendicular to OA, and qA is on the plane COA, and perpendicular to OA, pAq is the angle between the planes, and .. is equal to the angle A of the triangle, also the angle pOq is the angle subtended by BC i.e. is the angle a. Hence (Plane Trig. Art. 37) COR. 1.-If a.b.c. represent the lengths of the sides of the triangle, and r the radius, |